Page 359 - Advanced Thermodynamics for Engineers, Second Edition
P. 359

16.2 FURTHER CONSIDERATIONS OF BASIC ENGINE CYCLES             349




                  For the air-standard cycle the thermal efficiency is only a function of compression ratio, r.
                  The imep can be evaluated from Eqn (16.8).
                  With r ¼ 7,
                                          1         1   10   43000         1
                                                          5
                                                                             bar
                                p ¼  1    k 1                               5
                                         r     0:287   373   15  ð1   1=rÞ  10
                                        0:5408   1   43000
                                                             ¼ 16:90 bar:
                                  ¼
                                    0:287   373   15  ð1   1=7Þ
                  The maximum cycle pressure, p 3 , can be evaluated by calculating around the cycle from point 1.
                  Working around the cycle
                                                 k         1:4
                                         p 2 ¼ p 1 ðrÞ ¼ 1  ð7Þ  ¼ 15:25 bar:
                  This can be considered to be the point at which combustion (or, more accurately, energy transfer)
               commenced, initiated by a spark in this case.
                                                k 1         0:4
                                       T 2 ¼ T 1 ðrÞ  ¼ 373   7  ¼ 812:36 K:
                  The temperature at 3 can be evaluated from
                                                           q 23
                                                  T 3 ¼ T 2 þ
                                                            c v
                  The heat addition, q 23 , per unit mass of air, is
                                                          Q 0
                                                           v
                                                    q 23 ¼
                                                          ε
               giving
                                                  Q 0            43000
                                                   v
                                        T 3 ¼ T 2 þ  ¼ 812:36 þ
                                                 εc v          15   0:715
                                           ¼ 4821:7K:
                  (Note: this temperature is much higher than that achieved in an actual engine cycle, and would
               result in too high a thermal load on the structure).
                  It is now possible to evaluate the maximum pressure

                                                        mRT 3
                                                   p 3 ¼
                                                         V 3
               but

                                                        p 1 V 1
                                                    m ¼
                                                        RT 1
               and hence
                                               p 1 V 1 RT 3  p 1 V 1 T 3  T 3
                                                               ¼ p 1 r
                                          p 3 ¼       ¼
                                                RT 1 V 3  T 1 V 3   T 1
                                                     4821:7
                                                            ¼ 90:49 bar:
                                             ¼ 1   7
                                                       373
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